ANALYTICAL GEOMETRY 3D

ANALYTICAL GEOMETRY UNIT- 5 1. Condition for the plane lx + my + nz = 0 to touch the quadric cone ax2 + by2 + cz2 + 2fyz+ 2gzx + 2hxy = 0. (B.Sc.1989)

Let (x1, y1, z1) be the point of contact.

The tangent plane at (x1, y1, z1) is

x(a x1+ hy1+ gz1) + y (hx1+ by1+ fz1) + z (gx1+ fy1+ cz1) = 0. This is identical with the plane lx + my + nz = 0. 푎푥 + ℎ푦 + 푔푧 ℎ푥 + 푏푦 + 푓푧 푔푥 + 푓푦 + 푐푧 1 1 1 = 1 1 1 = 1 1 1 푙 푚 푛 If each ratio is k. ax1 + hy1 + gz1 - kl = 0 …(1) hx1 + by1 + fz1- km = 0 …(2) gx1 + fy1 + cz1- kn = 0 ... (3)

Since (x1, yı, z1) lies on lx + my + nz = 0

Therefore lx1 + my1 + nz1 = 0 ………(4)

Eliminating x1,y1, z1 from equations (1), (2), (3) and (4),we get 푎 ℎ 푔 푙 푓 푚 | ℎ 푏 |= 0. 푔 푓 푐 푛 푙 푚 푛 0 Simplifying, we get Al2 + Bm2 + Cn²+ 2F mn + 2G nl + 2H Im = 0 ... (5) where A, B, C, F, G, H are the cofactors of a, b, c, f, g, h in the 푎 ℎ 푔 determinant |ℎ 푏 푓| 푔 푓 푐 Multiplying (1) by A, (2) by H and (3) by G and adding, we get

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ANALYTICAL GEOMETRY 3D

∆x1 = k (Al+ Hm+ Gn) since ∆ = Aa + Hh + Gg, 0 = Ah + Hb+ Gf, 0= Ag + Hf+ Gc.

Similarly, ∆y1=k (Hl+ Bm+ Fn).

∆z1 = k (Gl+ Fm+ Cn). Hence the point of contact is given by the equations 푥 푦 푧 1 = 1 = 1 퐴푙 + 퐻푚 + 퐺푛 H푙 + Bm + Fn G푙 + Fm + Cn 푥 푦 푧 From condition (5), it can be seen that = = 푙 푚 푛 which is perpendicular to the plane lx + my + nz = 0 at the origin, is a generator of the cone Ax2+ By2 + Cz2 + 2F yz + 2G zx + 2H xy = 0…… (6) 퐴 퐻 퐺 In the determinant ∆' = |퐻 퐵 퐹|, we get 퐺 퐹 퐶 A' = BC- F2 = a ∆', F'= GH - AF = f∆' B' = CA - G2 = b∆', G' = HF- BG = g ∆' G' = AB - H2 = c∆', H' = FG – CH = h∆' Hence the perpendiculars to the tangent planes to the cone (6) generate the cone A'x2 + B'y2 + C'z2 + 2F'yz + 2G'zx + 2H'xy = 0 i.e., ax2+ by2 + cz2 + 2fyz + 2gzx + 2hxy = 0 The cones (6) and (7) are said to be reciprocal.

Example 1. Find the equations of the tangent planes to the 풙 풚 풛 cone 9x2 - 4y2+ 16z2 = 0 which contain the line = = ퟑퟐ ퟕퟐ ퟐퟕ The line is the intersection of the planes 72x - 32y = 0, i.e., 9x- 4y= 0. and 27y - 72z = 0, i.e., 3y- 8z = 0.

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ANALYTICAL GEOMETRY 3D

Hence any plane passing through this line is of the form 9x-4y+ 휆(3y- 8z) 0 9 ie., 9x+ y (3 휆 -4)-8 휆 z = 0. ... (1) This line touches the cone 9x2 - 4y2+ 16z2 = 0 …. (2) Hence the normal to the plane 풙 − 풙 풚 − 풚 풛 − 풛 ퟏ = ퟏ = ퟏ 풍 풎 풏 푥 푦 푧 = = 9 3 휆 − 4 −8 휆 is a generator of the reciprocal cone of the cone (2). Equation of the reciprocal cone of (2) is 푥2 푦2 푧2 − + = 0 … … (4) 9 4 16 (3) is a generator of cone (4). 92 (3 휆 − 4)2 (−8 휆)2 − + = 0 9 4 16 9휆2 − 24휆 + 16 64휆2 9 − + = 0 4 16 9휆2 − 24휆 + 16 9 − +4휆2 = 0 4 36 − 9휆2 + 24휆 − 16 + 16휆2 = 0 4 7휆2 + 24휆 + 20 = 0

Simplifying, we get 7휆2 + 24휆 + 20 = 0 −10 ie., 휆 = −2 표푟 ⁄7 Hence the equations of the planes are 휆 = −2, (1) gives 9x-10y + 16z = 0

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ANALYTICAL GEOMETRY 3D

−10 and 휆 = ⁄7, (1) gives 63x - 58y + 80z = 0.

Example 2. Find the general equation to a cone which touches the co-ordinate planes. If the co-ordinate planes touch a cone, the perpendiculars to co-ordinate planes touch the reciprocal cone. Hence the cone touching the co-ordinate planes is reciprocal to the cone passing through the c0-ordinate axes. The direction cosines of the co-ordinate axes are 1,0,0; 0,1,0; 0,0,1. The equation of the cone passing through the axis is of the form 2fyz + 2gzx+ 2hxy = 0. The required cone is the reciprocal cone of this cone and its equation is f2x2+g2y2+h2z2-2ghyz- 2hfzx- 2fgxy= 0.

This equation can be put in the form √푓푥+ √푔푦 + √ℎ푧 = 0.

2. The angle between the lines in which the plane ux + vy + wz = 0 cuts the cone. f(x,y,z)≡ ax2 +by2+ cz2+ 2fyz + 2gzx + 2hxy = 0. The plane meets the cone in two lines which pass through the origin and so 푥 푦 푧 the equations of the lines are of the form = = . 푙 푚 푛 The line lies in the plane and in the cone. Therefore, ul + vm + wn = 0 ... (1) and al2 +bm2+ cn2+ 2fyz + 2fmn + 2gnl + 2hlm = 0 ... (2) Eliminating n between (1) and (2), we get ul + vm (1) ⇒ 푛 = 푤

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ANALYTICAL GEOMETRY 3D

푢푙 + 푣푚 푢푙 + 푣푚 (2) ⇒ 푎푙2 + 푏푚2 + 푐(− )2 + 2푓푦푧 + 2푓푚 (− ) + 푤 푤 푢푙 + 푣푚 2푔 (− ) 푙 + 2ℎ푙푚 = 0 푤

푢2푙2 + 2푢푣푙푚 + 푣2푚2 2푓푢푙푚 + 2푓푣푚2 ⇒ 푎푙2 + 푏푚2 + 푐 ( ) − ( ) 푤2 푤 2푔푢푙2 + 2푔푣푙푚 − ( ) + 2ℎ푙푚 = 0 푤

⇒ 푎푙2푤2 + 푏푚2푤2 + 푐푢2푙2 + 2푐푢푣푙푚 + 푐푣2푚2 + 2푓푦푧푤2 − 2푓푢푤푙푚 − 2푓푣푤푚2 − 2푔푢푤푙2 − 2푔푣푤푙푚 + 2ℎ푙푚푤2 = 0

⇒ 푙2(푐푢2 + 푎푤2 − 2푔푢푤) + 2푙푚(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤) + 푚2(푐푣2 + 푏푤2 − 2푓푣푤) = 0 … (3)

The direction cosines of the two lines satisfy the equation (3) and if they are l1, m1, n1 and l2, m2, n2, we have 푙 −푏 ± √푏2 − 4푎푐 = 푚 2푎

푙 푚

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 −(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤) ± √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) = 2(푐푢2 + 푎푤2 − 2푔푢푤)

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ANALYTICAL GEOMETRY 3D

푙1

푚1

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 −(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤) + √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) = 2(푐푢2 + 푎푤2 − 2푔푢푤)

푙2

푚2

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 −(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤) − √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) = 2(푐푢2 + 푎푤2 − 2푔푢푤)

푙 푙 1 − 2 푚1 푚2

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 −(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤) + √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) = 2(푐푢2 + 푎푤2 − 2푔푢푤)

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 −(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤) − √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) − 2(푐푢2 + 푎푤2 − 2푔푢푤)

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 2√ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) 푙1푚2 − 푚1푙2 = 2 2 푚1푚2 2(푐푢 + 푎푤 − 2푔푢푤)

(ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤) 푙1푚2 − 푚1푙2 = 2 2 푚1푚2 (푐푢 + 푎푤 − 2푔푢푤)

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ANALYTICAL GEOMETRY 3D

푙 푚 − 푚 푙 푚 푚 1 2 1 2 = 1 2 (푐푢2 + 푎푤2 − 2푔푢푤) (ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤)

Sum of the roots 2 푙1 푙2 2(ℎ푤 + 푐푢푣 − 푓푤푣 − 푔푣푤) + = − 2 2 푚1 푚2 푐푢 + 푎푤 − 2푔푢푤

2 푙1푚2 + 푙2푚1 2(ℎ푤 + 푐푢푣 − 푓푤푣 − 푔푣푤) = − 2 2 푚1푚2 푐푢 + 푎푤 − 2푔푢푤

푙 푚 + 푙 푚 푚 푚 1 2 2 1 = 1 2 −2(ℎ푤2 + 푐푢푣 − 푓푤푣 − 푔푣푤) 푐푢2 + 푎푤2 − 2푔푢푤

Product of the roots

2 2 푙1 푙2 푐푣 + 푏푤 − 2푓푣푤 = 2 2 푚1 푚2 푐푢 + 푎푤 − 2푔푢푤

푙 푙 푚 푚 푙 푚 + 푙 푚 1 2 = 1 2 = 1 2 2 1 푐푣2 + 푏푤2 − 2푓푣푤 푐푢2 + 푎푤2 − 2푔푢푤 −2(ℎ푤2 + 푐푢푣 − 푓푤푣 − 푔푣푤)

푙 푚 − 푙 푚 = 1 2 2 1 (ℎ푤2 + 푐푢푣 − 푓푢푤 − 푔푣푤)2 √ −4(푐푢2 + 푎푤2 − 2푔푢푤)(푐푣2 + 푏푤2 − 2푓푣푤)

푙 푚 − 푙 푚 = 1 2 2 1 √±2푤(−퐴푢2 − 퐵푣2 − 퐶푤2 − 2퐹푣푤 − 2퐺푤푢 − 2퐻푢푣)

푙 푙 푚 푚 푙 푚 − 푙 푚 1 2 = 1 2 = 1 2 2 1 … … (4) 푐푣2 + 푏푤2 − 2푓푣푤 푐푢2 + 푎푤2 − 2푔푢푤 ±2푤푃

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ANALYTICAL GEOMETRY 3D

Where 푃 = −(퐴푢2 + 퐵푣2 + 퐶푤2 + 2퐹푣푤 + 2퐺푤푢 + 2퐻푢푣) and A, B, C, F, G, 푎 ℎ 푔 H are the cofactors of a, b, c, f, g, h in the determinant |ℎ 푏 푓|. 푔 푓 푐 From the symmetry, we get the expression in (4) is equal to

푛 푛 푚 푛 − 푚 푛 푛 푙 − 푙 푛 1 2 = 1 2 2 1 = 1 2 1 2 … … (5) 푎푣2 + 푏푢2 − 2ℎ푢푣 ±2푢푃 ±2푣푃

Each expression in (4) and (5) is

푙 푙 + 푚 푚 + 푛 푛 1 2 1 2 1 2 푐푣2 + 푏푤2 − 2푓푣푤 + 푐푢2 + 푎푤2 − 2푔푢푤 + 푎푣2 + 푏푢2 − 2ℎ푢푣

√∑(푚 푛 − 푚 푛 )2 = 1 2 2 1 ±2√(푢2 + 푣2 + 푤2)푃

If θ is the angle between the lines

푐표푠휃 푠𝑖푛휃 = 푙 푙 + 푚 푚 + 푛 푛 2 1 2 1 2 1 2 √∑(푚1푛2 − 푚2푛1)

푐표푠휃 푠𝑖푛휃 = … (6) (푎 + 푏 + 푐)(푢2 + 푣2 + 푤2) − 푓(푢, 푣, 푤) ±2√(푢2 + 푣2 + 푤2)푃

3. Condition that the cone has three mutually perpendicular generators. The condition that the plane should cut the cone in perpendicular generators is that θ = 90°. In that case by (6) of the previous article (a+ b+ c) (u² + v² + w2) = f (u, v, w). The third generator is perpendicular to these two generators. Hence it is normal to the plane containing these perpendicular generators. If the normal to the plane ux + vy+ wz = 0 lies on the cone, we have f (u, v, w) = 0.

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ANALYTICAL GEOMETRY 3D

∴ a+ b+ c = 0.

Example. Find the equation to the cone through the co-ordinate axes and the lines in which the plane lx + my+ nz = 0 cuts ax2 + by2 + cz2 + 2fyz+ 2gzx + 2hxy = 0. Let the equation of the cone passing through the co-ordinates axes by Fyz + Gzx+ Hxy = 0. Eliminating z between lx+ my + nz = 0 and ax2 + by2 + cz2 + 2fyz+ 2gzx + 2hxy = 0, we get −(lx + my) 푧 = 푛 −(lx+ my) −(lx+ my) −(lx+ my) ⇒ ax2 + by2 + c( )2+ 2fy( )+ 2g( )x + 2hxy = 0 푛 푛 푛 푙2푥2+푚2푦2+2푙푚푥푦 2푓푙푥푦 2푓푚푦2 2푔푙푥2 2푔푚푥푦 ⇒ ax2 + by2 + c( ) − − − − + 2ℎ푥푦 = 0 푛2 푛 푛 푛 푛 ⇒ a푛2푥2 + b푛2푦2 + c(푙2푥2 + 푚2푦2 + 2푙푚푥푦) − 2푓푙푛푥푦 − 2푓푚푛푦2 − 2푔푙푛푥2 − 2푔푚푛푥푦 + 2ℎ푛푥푦 = 0 ⇒ 푥2(a푛2 + 푐푙2 − 2푔푙푛) + 푥푦(2푐푙푚푥푦 − 2푓푙푛 − 2푔푚푛 + 2ℎ푛) + 푦2(b푛2 + 푐푚2 − 2푓푚푛) = 0 Similarly eliminate z between lx+ my + nz = 0 and Fyz + Gzx+ Hxy = 0.

−(lx+ my) −(lx+ my) Fy( ) + G( x+ Hxy = 0. 푛 푛 −퐹푙푥푦 − 퐹푚푦2 − 퐺푙푥2 − 퐺푚푥푦 + 퐻푛푥푦 = 0 퐹푙푥푦 + 퐹푚푦2 + 퐺푙푥2 + 퐺푚푥푦 − 퐻푛푥푦 = 0 퐺푙푥2 + (퐹푙 + 퐺푚 − 퐻푛)푥푦 + 퐹푚푦2 = 0 Since the two cones have common generators, we get a푛2 + 푐푙2 − 2푔푙푛 b푛2 + 푐푚2 − 2푓푚푛 = 퐺푙 퐹푚 Similarly eliminating x, we get the condition

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ANALYTICAL GEOMETRY 3D

a푛2 + 푐푙2 − 2푔푛푙 b푙2 + 푎푚2 − 2ℎ푚푙 = 퐺푛 퐻푚 Dividing by l on both sides, a푛2 + 푐푙2 − 2푔푙푛 b푙2 + 푎푚2 − 2ℎ푚푙 = 퐺푛푙 퐻푙푚

b푛2 + 푐푚2 − 2푓푚푛 = 퐹푚푛

Hence the equation of the required cone is

푙(b푛2 + 푐푚2 − 2푓푚푛)푦푧 + 푚(a푛2 + 푐푙2 − 2푔푙푛)푧푥 + 푛(b푙2 + 푎푚2 − 2ℎ푚푙)푥푦 = 0.

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